求和1*4+2*5+3*6+…+n*(n+1)

将 n(n+3) 拆成n^2+3n,再重新组成两个数列 :1^2,2^2,3^2,...,n^2和3,6,9,...,3n,利用公式即可求出解:
Sn=1×4+2×5+3×6+…+n(n+3)
=1×(1+3)+2(2+3)+3(3+3)+…+n^2+3n
=1^2+3+2^2+2×3+3^2+3×3+…+n^2+3*n
=(1^2+2^2+3^3+…+n^2)+3(1+2+3+…+n)
=n(n+1)(2n+1)/6+3n(n+1)/2
=n(n+1)[(2n+1)+9]/6
=n(n+1)(n+5)/3